$$ 1 \leftarrow $$ Claim. Fewer than $$ 9.5 \% $$ of homes have only a landline telephone and no wireless phone Sample data. A survey by the National Center for Health Statistics showed that among 14,473 homes $$ 5.83 \% $$ had landline

Question

UpStudy AI · Accepted Answer

To analyze the claim that fewer than $$ 9.5\% $$ of homes have only a landline telephone and no wireless phone, we can set up a hypothesis test.

### Step 1: Define the Hypotheses
- Null Hypothesis ($$ H_0 $$): $$ p \geq 0.095 $$ (The proportion of homes with only a landline is at least $$ 9.5\% $$)
- Alternative Hypothesis ($$ H_a $$): $$ p < 0.095 $$ (The proportion of homes with only a landline is less than $$ 9.5\% $$)

### Step 2: Gather Sample Data
From the survey:
- Sample size ($$ n $$) = 14,473
- Sample proportion ($$ \hat{p} $$) = $$ 5.83\% = 0.0583 $$

### Step 3: Calculate the Test Statistic
We can use the formula for the test statistic for a proportion:

$$
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
$$

Where:
- $$ p_0 = 0.095 $$ (the hypothesized population proportion)
- $$ \hat{p} = 0.0583 $$ (the sample proportion)
- $$ n = 14,473 $$ (the sample size)

### Step 4: Calculate the Test Statistic
Let's calculate the test statistic $$ z $$.

First, we need to calculate the standard error:

$$
SE = \sqrt{\frac{p_0(1 - p_0)}{n}}
$$

Now, we can compute both the standard error and the test statistic.

1. Calculate the standard error.
2. Calculate the test statistic $$ z $$.

Let's perform these calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$\sqrt{\frac{\left(0.095\left(1-0.095\right)\right)}{14473}}$$
- step1: Remove the parentheses:
  $$\sqrt{\frac{0.095\left(1-0.095\right)}{14473}}$$
- step2: Subtract the numbers:
  $$\sqrt{\frac{0.095\times 0.905}{14473}}$$
- step3: Multiply the numbers:
  $$\sqrt{\frac{0.085975}{14473}}$$
- step4: Divide the terms:
  $$\sqrt{\frac{3439}{578920000}}$$
- step5: Use the properties of radicals:
  $$\frac{\sqrt{3439}}{\sqrt{578920000}}$$
- step6: Simplify the expression:
  $$\frac{\sqrt{3439}}{200\sqrt{14473}}$$
- step7: Multiply by the Conjugate:
  $$\frac{\sqrt{3439}\times \sqrt{14473}}{200\sqrt{14473}\times \sqrt{14473}}$$
- step8: Multiply the numbers:
  $$\frac{\sqrt{49772647}}{200\sqrt{14473}\times \sqrt{14473}}$$
- step9: Multiply the numbers:
  $$\frac{\sqrt{49772647}}{2894600}$$
Calculate or simplify the expression $$ (0.0583 - 0.095) / \sqrt((0.095 * (1 - 0.095)) / 14473) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(0.0583-0.095\right)}{\sqrt{\frac{\left(0.095\left(1-0.095\right)\right)}{14473}}}$$
- step1: Remove the parentheses:
  $$\frac{0.0583-0.095}{\sqrt{\frac{0.095\left(1-0.095\right)}{14473}}}$$
- step2: Subtract the numbers:
  $$\frac{0.0583-0.095}{\sqrt{\frac{0.095\times 0.905}{14473}}}$$
- step3: Multiply the numbers:
  $$\frac{0.0583-0.095}{\sqrt{\frac{0.085975}{14473}}}$$
- step4: Divide the terms:
  $$\frac{0.0583-0.095}{\sqrt{\frac{3439}{578920000}}}$$
- step5: Simplify the root:
  $$\frac{0.0583-0.095}{\frac{\sqrt{49772647}}{2894600}}$$
- step6: Subtract the numbers:
  $$\frac{-0.0367}{\frac{\sqrt{49772647}}{2894600}}$$
- step7: Convert the expressions:
  $$\frac{-\frac{367}{10000}}{\frac{\sqrt{49772647}}{2894600}}$$
- step8: Multiply by the reciprocal:
  $$-\frac{367}{10000}\times \frac{2894600}{\sqrt{49772647}}$$
- step9: Reduce the numbers:
  $$-\frac{367}{50}\times \frac{14473}{\sqrt{49772647}}$$
- step10: Multiply the fractions:
  $$-\frac{367\times 14473}{50\sqrt{49772647}}$$
- step11: Multiply:
  $$-\frac{5311591}{50\sqrt{49772647}}$$
- step12: Multiply by the Conjugate:
  $$\frac{-5311591\sqrt{49772647}}{50\sqrt{49772647}\times \sqrt{49772647}}$$
- step13: Multiply the numbers:
  $$\frac{-5311591\sqrt{49772647}}{2488632350}$$
- step14: Reduce the fraction:
  $$\frac{-367\sqrt{49772647}}{171950}$$
- step15: Calculate:
  $$-\frac{367\sqrt{49772647}}{171950}$$
### Step 4: Calculate the Test Statistic

1. **Calculate the Standard Error (SE)**:
   $$
   SE \approx 0.002437
   $$

2. **Calculate the Test Statistic $$ z $$**:
   $$
   z \approx -15.057721
   $$

### Step 5: Determine the p-value
Since this is a one-tailed test, we will find the p-value corresponding to the calculated $$ z $$-value. A $$ z $$-value of $$ -15.057721 $$ is extremely low, indicating that the sample proportion is significantly less than the hypothesized proportion.

### Step 6: Conclusion
Given that the p-value will be very close to 0 (much less than any common significance level such as $$ 0.05 $$ or $$ 0.01 $$), we reject the null hypothesis.

### Final Conclusion
There is sufficient evidence to support the claim that fewer than $$ 9.5\% $$ of homes have only a landline telephone and no wireless phone.
There is sufficient evidence to support the claim that fewer than $$ 9.5\% $$ of homes have only a landline telephone and no wireless phone.

Claim. Fewer than of homes have only a landline telephone and no wireless phone Sample data. A survey by the National Center for Health Statistics showed that among 14,473 homes had landline

Solución de inteligencia artificial de Upstudy

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Statistics Questions